LXXII.4 (1995)
Class numbers of certain real abelian fields
by
Jae Moon Kim (Inchon)
0. Introduction. We fix an odd prime q. Let p be an odd prime such that p ≡ 1 mod q. Let
g(t) = (1 − t)
p−2+ 1
2 (1 − t)
p−3+ . . . + 1 p − 1 .
We will consider g(t) as an element in F
p[t], where F
pis the finite field with p elements. If necessary, we will also view g(t) as a polynomial in Z
p[t], where Z
pis the ring of p-adic integers. It is not hard to see that
tg(1 − t) ≡ (1 − t)
p− (1 − t
p)
p mod p.
This polynomial
f (t) = (1 − t)
p− (1 − t
p) p
in F
p[t] was first introduced by D. Mirimanoff around 1905 and has been exhaustively studied since then. For instance, he used the polynomial f (t) to prove the following striking criterion of A. Wieferich: if the Fermat quotient (2
p−1− 1)/p is not congruent to 0 mod p, then the first case of the Fermat’s Last Theorem is true (see [8]).
In this paper, we will study class numbers of certain real abelian fields by using the polynomial g(t). Our work is based on the observation that g(t) comes from a Coates–Wiles series. To be precise, let
h
t(x) = Y
w∈R
((1 + x)
w− t),
where R = {w ∈ Z
p| w
p−1= 1} is the group of the (p − 1)th roots of 1 in Z
p. Viewing h
t(x) as an element of Z
p[t][[x]], i.e., as a power series in x
This work was partially supported by the Basic Science Research Institute program, Ministry of Education,
#BSRI-94-1414.
[335]
with coefficients in Z
p[t], we have the following expansion (see [6]):
h
t(x) = (1 − t)
p−1+ g
1(t)x
p−1+ (higher terms) with
g
1(t) ≡ g(t) mod p.
To see why h
t(x) is a Coates–Wiles series, let t = s be a root of 1 in Z
p. Then h
s(x) is indeed a Coates–Wiles series (see [1], [10]).
In Section 1 of this paper we will use the above expansion of h
t(x) to fac- torize certain principal ideals into a product of prime ideals. In Section 2, we discuss class numbers of certain real abelian fields. Before we state the main theorems, we first explain several notations that will be used throughout this paper.
For each integer n ≥ 1, we choose a primitive nth root ζ
nof 1 so that ζ
mm/n= ζ
nwhenever n | m. Let k
0= Q(ζ
p), k
n= Q(ζ
pn+1), K
0= Q(ζ
pq) and K
n= Q(ζ
pn+1q). We denote the unique subfield of k
nof degree p
nover Q by Q
n. Let F
0= Q(ζ
q), F
n= Q
n(ζ
q), F
0+= Q(ζ
q+ ζ
q−1) and F
n+= Q
n(ζ
q+ ζ
q−1). Thus for E = k, K, F and F
+, E
nis the nth layer of the basic Z
p-extension of E
0. We denote the Galois groups Gal(F
0/Q) and Gal(F
0+/Q) by ∆ and ∆
+, respectively, and use the same letters ∆, ∆
+for those Galois groups isomorphic to Gal(F
0/Q), Gal(F
0+/Q). Elements of ∆
+and ∆ will be arranged as ∆
+= {τ
1, τ
2, . . . , τ
l= id} and ∆ = {±τ
1, ±τ
2, . . . , ±τ
l} with l =
12ϕ(q). For each τ ∈ ∆, let p(τ ) be the integer modulo q corresponding to τ under the natural isomorphism ∆ ' (Z/qZ)
×. Note that p(−τ ) = −p(τ ).
Finally, we let σ be the topological generator of Γ = lim ←− Gal(K
n/K
0) which maps ζ
pnto ζ
p1+pnfor each n ≥ 1 and ζ
qto ζ
q. Restrictions of σ to various subfields k
n, F
n, F
n+and K
nof K
∞= S
n≥0
K
nare also denoted by σ.
Now we state the main theorems of this paper:
Theorem 2. If p divides Q
χ∈∆
b
+,χ6=1B
1,χω−1, then p divides the class number of F
n+for all n ≥ 1, where ω is the Teichm¨uller character on Gal(k
0/Q) ' (Z/pZ)
×.
Theorem 3. If p does not divide Q
χ∈∆
b
+,χ6=1B
1,χω−1, then the prime ideals of F
n+above p are of order prime to p in the ideal class group of F
n+. It is well known, by the class number formula, that the relative class number h
−K0
of the field K
0is given by the formula h
−K0
= Qw Q
%
−
12B
1,%, where the product is taken over all odd characters of Gal(K
0/Q). Hence Q
χ∈∆
b
+,χ6=1B
1,χω−1contributes to h
−K0
. Thus from Theorems 2 and 3 we obtain some information on the p-divisibility of h
F+n
= h
+Fn
, the plus part of the class number of F
n, from that of the minus part h
−K0
.
In Section 3, we prove a lemma to finish the proof of Theorem 2. This lemma treats certain relations among cyclotomic units. For a deeper analysis of the relations of cyclotomic units, we refer to [2]. We will apply the results of [2] to our special situation. A similar, but slightly different computation was performed in [7].
1. Factorization of a certain principal ideal. In this section, we start out with an explicit element ξ
nin F
n+whose norm to F
0+equals 1. So, by the Hilbert Theorem 90, ξ
nis of the form ξ
n= α
σ−1nfor some α
n∈ F
n+. The aim of this section is to factorize the principal ideal (α
n) into a product of prime ideals of F
n+. This factorization is crucial in the proofs of Theorems 2 and 3 of the following section.
Let
ξ
n= Y
w∈R
(ζ
pwn+1− ζ
q)(ζ
pwn+1− ζ
q−1).
Then ξ
nis an element of F
n+since ξ
n= N
Kn/Fn+
(ζ
pn+1− ζ
q). Since each ζ
pwn+1− ζ
q±1is a cyclotomic unit in K
n, so is ξ
n. Thus ξ
ncan be thought of as a cyclotomic unit in F
n+in the sense of W. Sinnott (see [9]). One can easily check that N
F+n/F0+
(ξ
n) = 1. Indeed, N
F+n/F0+
(ξ
n) = N
Kn/F0+
(ζ
pn+1− ζ
q) = N
K0/F0+
(N
Kn/K0(ζ
pn+1− ζ
q))
= N
K0/F0+
(ζ
p− ζ
q) = N
F0/F0+
Y
1≤i≤p−1
ζ
pi− ζ
q= N
F0/F0+
1 − ζ
qp1 − ζ
q= 1.
The last equality holds since p ≡ 1 mod q. Hence ξ
n= α
σ−1nfor some α
n∈ F
n+by the Hilbert Theorem 90.
Lemma 1. ξ
n= α
σ−1nfor some p-unit α
n∈ F
n+. P r o o f. Let K
∞= S
n≥0
Q(ζ
pn+1q) be the basic Z
p-extension of K
0. Let E
∞0(C
∞) be the group of p-units (cyclotomic units) of K
∞. In [4], Iwa- sawa proves that the cohomology group H
1(Γ, E
∞0) is a finite group, where Γ = Gal(K
∞/K
0). But H
1(Γ, C
∞) ' (Q
p/Z
p)
l, where l =
12ϕ(q) (see [5]).
Thus the induced map H
1(Γ, C
∞) → H
1(Γ, E
∞0) from the natural inclusion C
∞→ E
∞0must be a zero map. Since the inflation maps on H
1are injec- tive, H
1(G
n, C
n) → H
1(G
n, E
n0) is a zero map, where G
n= Gal(K
n/K
0) and C
n(E
n0) is the group of cyclotomic units (p-units) in K
n. Then by the cyclicity of the group G
n, H
−1(G
n, C
n) → H
−1(G
n, E
n0) is also a zero map, which means that a cyclotomic unit in K
nwhose norm to K
0equals 1 is of the form β
σ−1for some p-unit β ∈ K
n. In particular, since
N
Kn/K0( Q
w∈R
(ζ
pwn+1− ζ
q)) = 1, we have Y
w∈R
(ζ
pwn+1− ζ
q) = β
σ−1for some p-unit β ∈ K
n. Thus
Y
w∈R
(ζ
pwn+1− ζ
q)
pn−1= N
Kn/FnY
w∈R
(ζ
pwn+1− ζ
q)
(pn−1)/(p−1)= N
Kn/Fn(β
(pn−1)/(p−1))
σ−1. Note that Q
w∈R
(ζ
pwn+1− ζ
q)
pn= u
σ−1for some cyclotomic unit u ∈ F
nsince p
nH
−1(G
n, C
Fn) = 0. Here C
Fnis the group of cyclotomic units of F
n. Therefore
Y
w∈R
(ζ
pwn+1− ζ
q) = (uN
Kn/Fn(β
−(pn−1)/(p−1)))
σ−1. Put α
n= N
Fn/Fn+
(uN
Kn/Fn(β
−(pn−1)/(p−1))). Then ξ
n= α
σ−1nand α
nis a p-unit in F
n+. This proves the lemma.
Fix a prime ideal ℘
0of F
0+above p and let ℘
nbe the prime ideal of F
n+above ℘
0. Then {℘
τ0i| τ
i∈ ∆
+} and {℘
τni| τ
i∈ ∆
+} are the sets of all prime ideals of F
0+and F
n+above p. For each ℘
τni, there are two prime ideals in F
nabove ℘
τni. By abuse of notation, we write them as ℘
τniand ℘
−τn i. This will not cause any confusion. Since primes of F
nabove p totally ramify in K
n, above each prime ideal ℘
±τn ithere is a unique prime ideal e ℘
±τn iin K
n. For each τ ∈ ∆ = {±τ
1, . . . , ±τ
l}, let F
n,℘τnbe the completion of F
nat ℘
τn, and let ϕ
τ: F
n→ F
n,℘τnbe the natural embedding. Put s
τ= ϕ
τ(ζ
q), which is a qth root of 1 in Z
p. For brevity, we write s for s
id= ϕ
id(ζ
q). Then s
p(τ )τ= ϕ
τ(ζ
q)
p(τ )= ϕ
τ(ζ
qp(τ )) = ϕ
τ(ζ
qτ). Since the completion of F
nτat ℘
τnis the same as the completion of F
nat ℘
n, we have s
p(τ )τ= ϕ
τ(ζ
qτ) = ϕ
id(ζ
q) = s.
Therefore s
τ= s
p(τ−1)and s
p(ττ 0)= s
p(τ−1)p(τ0)= s
p(τ−1τ0)for any τ, τ
0∈ ∆.
Theorem 1. Let ξ
nbe as before and write ξ
n= α
σ−1nfor some p-unit α
n∈ F
n+as in Lemma 1. If (α
n) = ℘
Σn1≤i≤laiτiis the factorization in F
n+, then a
i≡ 2g(s
p(τi−1)) mod p, where g(t) is the polynomial defined in the introduction.
R e m a r k. If ξ
n= α
σ−1n= (α
0n)
σ−1for another p-unit α
0n, then α
n= α
n0α
0for some p-unit α
0in F
0+. Since the primes of F
0+above p totally ramify in F
n+, their ramification indices are p
n. Hence a
i’s are uniquely determined mod p
n, so mod p.
One can show that g(s) = g(s
−1) for any (p − 1)th root s in Z
p(see [6]).
Thus
g(s
p(−τi−1)) = g(s
−p(τi−1)) = g(s
p(τi−1)), and g(s
p(τi−1)) is well defined for each τ
i∈ ∆
+.
P r o o f o f T h e o r e m 1. To compute a
i, we read the prime factoriza- tion (α
n) = ℘
Σan iτiin the field K
n, ˜℘τin
(or in K
n, ˜℘−τin
). Since π
n= ζ
pn+1− 1 generates the prime ideal of K
n, ˜℘τin
, we have (α
n) = ℘
Σan iτi= (π
n)
ai. Hence α
n= π
aniη for some unit η in K
n, ˜℘τn. Note that η
σ−1≡ 1 mod (π
np). We claim that π
nσ−1≡ 1 + π
p−1nmod (π
pn). First, notice that for each 1 ≤ k ≤ p − 1, ζ
p1+kpn+1− 1 = ζ
pn+1ζ
pkn− 1 ≡ ζ
pn+1− 1 mod (ζ
pn− 1). Thus
Y
1≤k≤p−1
(ζ
p1+kpn+1− 1) ≡ Y
1≤k≤p−1
(ζ
pn+1− 1)
= (ζ
pn+1− 1)
p−1mod (ζ
pn− 1).
Therefore
π
nσ−1= ζ
p1+pn+1− 1
ζ
pn+1− 1 = ζ
pn+ ζ
pn− 1 ζ
pn+1− 1
= ζ
pn+ Y
1≤k≤p−1
(ζ
p1+kpn+1− 1)
≡ 1 + (ζ
pn+1− 1)
p−1mod (ζ
pn− 1) as claimed. Hence
(π
naiη)
σ−1≡ (1 + π
np−1)
ai≡ 1 + a
iπ
np−1mod (π
np).
On the other hand, by putting x = ζ
pn+1−1 and t = s
τin Q
w∈R
((1 + x)
w− t) = (1 − t)
p−1+ g
1(t)x
p−1+ (higher terms), we obtain Y
w∈R
(ζ
pwn+1− s
τ) ≡ 1 + g(s
τ)π
np−1mod (π
np).
Hence if we view ξ
n= Q
w∈R
(ζ
pwn+1− ζ
q)(ζ
pwn+1− ζ
q−1) as an element of K
n, ˜℘τn, we have
ξ
n= Y
w∈R
(ζ
pwn+1− s
τ)(ζ
pwn+1− s
−1τ)
≡ (1 + g(s
τ)π
np−1)(1 + g(s
−1τ)π
p−1n)
≡ 1 + (g(s
τ) + g(s
−1τ))π
np−1mod (π
np).
Thus 1 + a
iπ
p−1n≡ α
σ−1n= ξ
n≡ 1 + (g(s
τi) + g(s
−1τi))π
np−1mod (π
np). There- fore a
i≡ g(s
τi) + g(s
−1τi) ≡ 2g(s
τi) ≡ 2g(s
p(τi−1)) mod (π
n), hence mod p.
2. Main theorems. Recall that l = [F
0+: Q] and that we arranged
elements of ∆
+as ∆
+= {τ
1, τ
2, . . . , τ
l−1, τ
l= id}. Let A
0= (a
ij) be the
l × l matrix with entries in F
psuch that a
ij≡ g(s
p(τi−1τj)) mod p. Note that each row and column of A
0has a fixed sum g = P
1≤i≤l
g(s
p(τi)). Let A = (b
ij) be the l × l matrix with entries in F
psuch that
b
ij=
a
ijif j ≤ l − 1, 1 if j = l.
Below, we write (a
1, . . . , a
n)
tfor the column vector with entries a
1, . . . , a
n. Lemma 2. Suppose det A ≡ 0 mod p. Then there exists b = (b
1, . . . , b
l)
tin F
lpsuch that
(i) A
0b ≡ 0 = (0, . . . , 0)
tmod p,
(ii) b is not a constant multiple of 1 = (1, . . . , 1)
t. P r o o f. We examine the following two cases separately.
C a s e 1: g 6≡ 0 mod p. By adding each column of A
0to the last one, we have
det A
0= det
a
ij j≤l−1g .. . g
= g det A ≡ 0 mod p.
Hence A
0b = 0 has a nontrivial solution. This solution cannot be a multiple of 1, for otherwise, the row sum g would be 0.
C a s e 2: g ≡ 0 mod p. Since each row sum is 0, 1 is obviously a solution of A
0b = 0. To prove the existence of a solution which is not a multiple of 1, it is enough to check that the F
p-rank of A
0is less than or equal to l − 2.
Let B be the (l − 1) × (l − 1) matrix consisting of the first (l − 1) × (l − 1) entries in A
0. By performing elementary row and column operations, we see that
rank A
0= rank
B
0 .. . 0 0 . . . 0
0
= rank B.
By adding each row of A to the last one, we have
det A = det
B
1 .. . 1 0 . . . 0
l
= l det B.
Since l 6≡ 0 mod p, det B ≡ 0 mod p. Therefore rank A
0= rank B ≤ l − 2.
Let ω be the Teichm¨ uller character on Gal(k
0/Q) ' (Z/pZ)
×and χ be a
character on ∆
+. For the proofs of the main theorems, we need the following
theorem which interprets det A in terms of the generalized Bernoulli numbers B
1,χω−1.
Theorem. det A ≡ 0 mod p if and only if Q
χ∈∆
b
+,χ6=1B
1,χω−1≡ 0 mod p.
P r o o f. See Theorem 3 of [7].
Now we restate and prove the main theorems.
Theorem 2. If p | Q
χ∈∆
b
+,χ6=1B
1,χω−1, then p | h
F+n
for all n ≥ 1.
P r o o f. Since h
F+ n| h
F+m
for all n ≤ m by the class field theory, it is enough to show that p | h
F+1
. By the above theorem, we can assume that det A ≡ 0 mod p. Then by Lemma 2, there exists a vector b = (b
1, . . . , b
l)
tin F
lpsatisfying those two conditions (i), (ii) in the lemma. Suppose p - h
F+1
. Then, by the class field theory, we have
(iii) p - h
F+ 0.
Moreover, the Sylow p-subgroup of E/C must be trivial (see [9]), where E(C) is the group of units (cyclotomic units) in F
1+. Thus, the cohomology groups H
i(G
1, E/C) are trivial for all i ∈ Z. Hence by considering the long exact sequence of cohomology groups coming from the short exact sequence 0 → C → E → E/C → 0, we have
(iv) The homomorphism H
−1(G
1, C) → H
−1(G
1, E) induced by the inclusion C → E is an isomorphism.
Let δ = ξ
Σli=1biτi= α
(Σbiτi)(σ−1)with b = (b
1, . . . , b
l)
tas before and ξ = ξ
1, α = α
1as in Theorem 1. Then the principal ideal (α
Σbiτi) factorizes as
(α
Σbiτi) = ℘
(2Σlj=1g(sp(τ −1 j )
)τj)(Σli=1biτi)
1
.
In this expression,
X
lj=1
g(s
p(τj−1))τ
jX
li=1
b
iτ
i= X
1≤i,j≤l
g(s
p(τj−1))b
iτ
jτ
i= X
l k=1X
τjτi,ji=τk
g(s
p(τj−1))b
iτ
k= X
l k=1X
li=1
g(s
p(τk−1τi))b
iτ
k.
Since A
0b ≡ 0, P
li=1
g(s
p(τk−1τi))b
i≡ 0 mod p for each k = 1, . . . , l. Hence
(α
Σbiτi) = ℘
Σ1li=1diτifor some d
isatisfying d
i≡ 0 mod p. Since ℘
p1= ℘
0, we get (α
Σbiτi) = I
0for some ideal I
0of F
0+. But the subgroup of the ideal class group of F
0+consisting of ideal classes that become principal in F
1+is a p-group. Thus nonprincipal ideals of F
0+cannot capitulate in F
1+by (iii). Therefore
(α
Σbiτi) = I
0= (α
0)
for some α
0∈ F
1+, and thus α
Σbiτi= α
0η
0for some unit η
0in F
1+. Then we have
δ = ξ
Σbiτi= α
(Σbiτi)(σ−1)= (η
0)
σ−1.
Since the induced homomorphism H
−1(G
1, C) → H
−1(G
1, E) is an isomor- phism by (iv), (η
0)
σ−1= η
σ−1for some cyclotomic unit in F
1+. However, this cannot happen by (ii) and the following lemma which will be proved in the next section.
Lemma 3. Let ξ = Q
w∈R
(ζ
pw2− ζ
q)(ζ
pw2− ζ
q−1) as before. If ξ
Σli=1ciτi= η
σ−1for some cyclotomic unit η ∈ F
1+, then c
1≡ . . . ≡ c
lmod p.
Theorem 3. If p - Q
χ∈∆
b
+,χ6=1B
1,χω−1, then the prime ideals of F
n+above p are of order prime to p in the ideal class group of F
n+for all n ≥ 0.
P r o o f. It is enough to show that the ideal class [℘
n] is of order prime to p. As in Lemma 2, we examine two cases separately.
C a s e 1: g 6≡ 0 mod p. Since det A
0= g det A, det A
06≡ 0 mod p. Thus there exists x = (x
1, . . . , x
l)
tsuch that A
0x = (0, . . . , 0, 1)
t. Let ξ
nand α
nbe as in Theorem 1. Then (α
Σnli=1xiτi) = ℘
Σnli=1diτifor some d
isatisfying d
l≡ 1 mod p and d
1≡ . . . ≡ d
l−1≡ 0 mod p. Since ℘
pn= ℘
n−1, we get (∗) (α
Σxn iτi) = ℘
nI
n−1for some ideal I
n−1of F
n−1+whose prime factors lie above p. Let mp
kbe the order of the ideal class [℘
n] with (m, p) = 1. If k 6= 0, then I
n−1mpk−1is a principal ideal in F
n+. Therefore by raising both sides of (∗) to the power of dp
k−1we get a contradiction. Hence k = 0.
C a s e 2: g ≡ 0 mod p. Since det A 6≡ 0 mod p, F
p-rank A
0= l − 1.
Thus columns of A
0except the last one are linearly independent over F
pand are contained in the subspace of F
lpconsisting of {y = (y
1, . . . , y
l)
t∈
F
lp| y
1+ . . . + y
l= 0}, which is of dimension l − 1. Therefore if we view
A
0as a linear map from F
lpto F
lp, the image of A
0is precisely the subspace
described above. For each i, 1 ≤ i ≤ l − 1, choose b
iin F
lpsuch that
A
0b
i= (0, . . . , 1, . . . , 0, −1)
t, with 1 at the ith place, −1 at the last place
and 0 elsewhere. Then as in the first case, we get (∗∗) ℘
τni−1I
n−1= (β
n)
for some β
n∈ F
n+. Note that ℘
Σnli=1τiis the prime ideal of Q
n, hence is principal. Thus by multiplying (∗∗) for 1 ≤ i ≤ l we see that ℘
lnI
n−10is a principal ideal for some ideal I
n−10whose prime factors lies above p. Since p - l, we can check that [℘
n] is of order prime to p as in Case 1.
Let A
nbe the Sylow p-subgroup of the ideal class group of F
n+and let A
∞= lim ←− A
n, where the limit is taken under the norm maps. It is well known that A
∞' Z
λp⊕M for some finite group M which measures the capitulation.
It is conjectured that the Iwasawa λ-invariant for F
∞+/F
0+equals 0 and R. Greenberg gave several equivalent statements to this (see [3]). By using one of the equivalent statements (Theorem 2 of [3]), we have the following corollary.
Corollary. Suppose λ = 0. Then p - Q
χ∈∆
b
+,χ6=1B
1,χω−1if and only if A
∞= {0}.
P r o o f. Theorem 2 takes care of the if part. For the converse, we need the assumption λ = 0. Since λ = 0, #A
nis bounded by #M as n → ∞.
Equivalently, every ideal class in A
Gnncontains an ideal whose prime factors lie above p by Theorem 2 of [3]. Thus if p - Q
χ∈∆
b
+,χ6=1B
1,χω−1, then A
Gnn= {0} by Theorem 3. Since A
nand G
nare p-groups, A
n= {0} for all n.
Therefore A
∞= {0}.
3. Proof of Lemma 3. In this section we prove Lemma 3 stated in the previous section. The proof is based on the work of V. Ennola ([2]) and is similar to that of Theorem 1 in [7]. In particular we need the following theorem:
Theorem (V. Ennola). Let δ = Q
1≤a<n
(1 − ζ
na)
xa, x
a∈ Z, be a cyclo- tomic unit in Q(ζ
n). For an even character χ 6= 1 of conductor f belonging to Q(ζ
n), define Y (χ, δ) by
Y (χ, δ) = X
f |d|nd
1
ϕ(d) T (χ, d, δ) Y
p|d
(1 − χ(p)), where
T (χ, d, δ) =
d−1
X
(a,d)=1a=1
χ(a)x
na/d.
If δ is a root of 1, then Y (χ, δ) = 0 for all even characters χ 6= 1 belonging
to Q(ζ
n).
We also need the following properties of Y which are easy to check. Let χ 6= 1 be an even character belonging to Q(ζ
n). Then
(i) Y (χ, δ
1δ
2) = Y (χ, δ
1) + Y (χ, δ
2).
(ii) If (root of 1)×δ
1= (root of 1)×δ
2, then Y (χ, δ
1) = Y (χ, δ
2).
(iii) For any σ ∈ Gal(Q(ζ
n)/Q), Y (χ, δ
σ) = χ(σ)Y (χ, δ).
(iv) Y (χ, δ
σ−1) = (χ(σ) − 1)Y (χ, δ).
Now we sketch the proof of Lemma 3 briefly.
S k e t c h o f p r o o f. Suppose ξ
Σli=1ciτi= η
σ−1, with ξ, η as in the lemma. By (ii),
Y (%, ξ
Σciτi) = Y (%, η
σ−1)
for any even character % 6= 1 in Gal(K
1/Q)
∧. By (i), (iv), we obtain (∗)
X
l i=1c
i%(τ
i)Y (%, ξ) = (%(σ) − 1)Y (%, η).
Fix a nontrivial character ψ of Gal(Q
1/Q). For a nontrivial character χ ∈
∆ b
+, put % = χψ in (∗). After a similar computation to that of Theorem 1 of [7], we have
(p − 1) X
l i=1c
iχ(τ
i) = (ψ(σ) − 1)α(χ)
for some algebraic integer α(χ). By letting χ run through all the nontriv- ial even characters of b ∆
+, we have the following system of linear equa- tions:
(p − 1)T (c
1, . . . , c
l)
t= (ψ(σ) − 1)(. . . , α(χ), . . .)
t,
where T is the (l − 1) × l matrix with rows of the form (χ(τ
1), . . . , χ(τ
l)).
Let L = Q(ζ
p, α(χ)’s, χ(τ
i)’s), O
Lbe the ring of integers of L, and P be a prime ideal of O
Labove p. Then we have
T (c
1, . . . , c
l)
t≡ (0, . . . , 0)
tmod P, since ψ(σ) − 1 ≡ ζ
p− 1 ≡ 0 mod P.
Let T be the matrix obtained by reducing the entries of T mod P. By
using Lemma 1.2 of [7], one can check that the O
L/P-rank of T is l−1. Hence
{x = (x
1, . . . , x
l)
t∈ O
L/P | T x = (0, . . . , 0)
t} is one-dimensional. But
(1, . . . , 1)
tobviously satisfies T (1, . . . , 1)
t= (0, . . . , 0)
t. Thus (c
1, . . . , c
l)
t≡
α(1, . . . , 1)
tmod P for some α ∈ O
L. Therefore c
1≡ . . . ≡ c
lmod P, hence
mod p.
References
[1] J. C o a t e s and A. W i l e s, On p-adic L-functions and elliptic units, J. Austral.
Math. Soc. 26 (1978), 1–25.
[2] V. E n n o l a, On relations between cyclotomic units, J. Number Theory 4 (1972), 236–247.
[3] R. G r e e n b e r g, On the Iwasawa invariants of totally real number fields, Amer. J.
Math. 98 (1976), 263–284.
[4] K. I w a s a w a, On cohomology groups of units for Z
p-extensions, ibid. 105 (1983), 189–200.
[5] J. M. K i m, Cohomology groups of cyclotomic units, J. Algebra 152 (1992), 514–519.
[6] —, Coates–Wiles series and Mirimanoff’s polynomial, J. Number Theory, to ap- pear.
[7] —, Units and cyclotomic units in Z
p-extensions, Nagoya Math. J. (1995), to appear.
[8] P. R i b e n b o i m, 13 Lectures on Fermat’s Last Theorem, Springer, New York, 1972.
[9] W. S i n n o t t, On the Stickelberger ideal and the circular units of a cyclotomic field, Ann. of Math. 108 (1978), 107–134.
[10] L. W a s h i n g t o n, Introduction to Cyclotomic Fields, Graduate Texts in Math. 83, Springer, New York, 1980.
DEPARTMENT OF MATHEMATICS INHA UNIVERSITY, INCHON, KOREA E-mail: JMKIM@MUNHAK.INHA.AC.KR